Re: Re: Proving inequalities with Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg52553] Re: [mg52499] Re: [mg52491] Proving inequalities with Mathematica*From*: DrBob <drbob at bigfoot.com>*Date*: Wed, 1 Dec 2004 05:59:21 -0500 (EST)*References*: <200411290622.BAA27997@smc.vnet.net> <200411301024.FAA01299@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

I wish I'd thought of that. Brilliant, Andrzej. FWIW, here's a process for making other examples (now that you've pointed the way): Clear[a,b] a[n_]=(100/99)^-n; Simplify[Abs[a[n+1]/a[n]]<1,n>1] True (If that fails, choose another series.) Now choose a[1] and a[2]: s = Simplify[Sum[a[n], {n, 3, Infinity}]]; {a[1], a[2]} = {x, y} /. First@FindInstance[{x + y == -s, Abs@x > Abs@y > Abs@a@3}, {x, y}, Reals] {-(490001/5000), 9703/10000} All premises are guaranteed at this point (setting b == a), but the conclusion can NEVER be true, since Sum[a[i]^2, {i, 1, Infinity}] < a[1]^2 is equivalent to: Sum[a[i]^2, {i, 2, Infinity}] < 0 But wait -- could FindInstance fail? Consider this, just for s >= 0: s =. ; Reduce[{s >= 0, x + y == -s, Abs[x] > Abs[y] > Abs[a3], a3 \[Element] Reals}, {x, y}, Reals] s > 0 && ((a3 <= -(s/2) && x < a3 - s) || (Inequality[-(s/2), Less, a3, LessEqual, 0] && (x < a3 - s || -a3 - s < x < -(s/2))) || (0 < a3 < s/2 && (x < -a3 - s || a3 - s < x < -(s/2))) || (a3 >= s/2 && x < -a3 - s)) && y == -s - x The Or in the middle is a run-down of all the ways a3 can stand in relation to +-s/2 and 0; for every case we have a non-empty range in which to choose x and a formula for y. The situation is similar for s<0. Bobby On Tue, 30 Nov 2004 05:24:01 -0500 (EST), Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > > On 29 Nov 2004, at 15:22, Toshiyuki (Toshi) Meshii wrote: > >> Hi, >> >> I was wondering whether Mathematica is useful for proving a problem of >> inequality. >> My problem is as follows: >> >> Let An and Bn (n=1,2,3...) be real sequences. >> Some characteristics of these sequences are known. >> >> i) Abs[An+1/An] < 1 >> ii) Abs[Bn+1/Bn] < 1 >> iii) Sum[An, {1,Infinity}] = 0 >> iv) Sum[An*Bn, {1, Infinity}] = alpha (note: a real number) >> >> Then I want to prove with Mathematica that >> 0 < Abs[alpha] < Abs[A1*B1] >> >> Does anyone have an idea? >> >> -Toshi >> > > First of all, you have written your condition as > >> Abs[An+1/An] < 1 > > Presumably you meant n+1 to be a subscript (if not the condition is > always false). > Assuming that, Mathematica can certainly be helpful in proving your > claim to be false. Here is an example. > > We define the sequence A[n] as follows: > > A[1] = -1; A[2] = Log[2]; > A[n_] = (-1)^(n - 1)/(n - 1); > > You can check that the condition Abs[A[n+1]/A[n]]<1 is always > satisfied. We also define the sequence B[n] by > B[i_] := A[i] > so that the condition > > Abs[B[n+1]/B[n]]<1 > > is satisifed. > > We check that (iii) is satisfied: > > > A[1]+A[2]+Sum[A[n],{n,3,Infinity}]//Simplify > > 0 > > Let's now compute alpha: > > > alpha=A[1]*B[1]+A[2]*B[2]+Sum[A[i]*B[i],{i,3,Infinity}]//N > > > 2.12539 > > Let's see if your inequality holds: > > > Abs[alpha]<Abs[A[1]*B[1]] > > False > > Oops ... > > > Andrzej Kozlowski > Chiba, Japan > http://www.akikoz.net/~andrzej/ > http://www.mimuw.edu.pl/~akoz/ > > > > -- DrBob at bigfoot.com www.eclecticdreams.net